Mathematics.

systems and stability

Systems of Differential Equations

Differential Equations35 minDifficulty7 out of 10

You should know: second order differential equation, matrices

Overview

A system of first-order linear differential equations describes several interacting quantities simultaneously, x′ = Ax, where x(t) is a vector of unknown functions and A is a constant coefficient matrix. Any higher-order single ODE can be rewritten as an equivalent first-order system by introducing derivatives as new variables, so systems are the general framework underlying all ODE theory. When A has a full set of eigenvectors, the solution is a combination of exponential modes eᵗλv, one per eigenvalue-eigenvector pair, exactly mirroring the characteristic-root approach for scalar equations. Systems naturally model coupled phenomena such as predator-prey populations, multi-compartment chemical reactions, and multi-mass spring networks.

Intuition

Think of a system x′=Ax as several quantities pushing and pulling on each other's rates of change, the way rabbit and fox populations each affect the other's growth rate. Diagonalizing (finding eigenvectors of) A finds the special combinations of the variables that DON'T interact — each such combination grows or decays purely exponentially at its own eigenvalue rate, decoupled from the rest. The general solution is then just a blend of these independent 'normal modes,' exactly as decomposing a vector into eigenbasis coordinates simplifies any linear transformation.

Formal Definition

Definition

For a linear system with constant coefficient matrix A, solutions are built from eigenvalues λ and eigenvectors v of A:

x′(t)=Ax(t)\mathbf{x}'(t) = A\mathbf{x}(t)
Linear constant-coefficient system
Av=λvA\mathbf{v} = \lambda \mathbf{v}
Eigenvalue problem
x(t)=C1eλ1tv1+C2eλ2tv2\mathbf{x}(t) = C_1 e^{\lambda_1 t}\mathbf{v}_1 + C_2 e^{\lambda_2 t}\mathbf{v}_2
General solution (2×2 case, distinct real eigenvalues)

Worked Examples

  1. Write the coefficient matrix and find its eigenvalues from det(A-λI)=0.

    A=(1221),(1−λ)2−4=0  ⟹  λ=3,−1A=\begin{pmatrix}1&2\\2&1\end{pmatrix}, \quad (1-\lambda)^2 - 4 = 0 \implies \lambda = 3, -1

  2. Find eigenvector for λ=3: (A-3I)v=0 gives -2v₁+2v₂=0, so v=(1,1).

    λ1=3, v1=(1,1)\lambda_1=3,\ \mathbf{v}_1=(1,1)

  3. Find eigenvector for λ=-1: (A+I)v=0 gives 2v₁+2v₂=0, so v=(1,-1).

    λ2=−1, v2=(1,−1)\lambda_2=-1,\ \mathbf{v}_2=(1,-1)

  4. General solution and apply initial conditions x(0)=4, y(0)=0.

    x(t)=C1e3t(1,1)+C2e−t(1,−1);C1+C2=4, C1−C2=0  ⟹  C1=C2=2\mathbf{x}(t) = C_1e^{3t}(1,1) + C_2e^{-t}(1,-1); \quad C_1+C_2=4,\ C_1-C_2=0 \implies C_1=C_2=2

Answer: x(t) = 2e^(3t) + 2e^(-t), y(t) = 2e^(3t) − 2e^(-t) (verify: x′=6e^{3t}-2e^{-t}; x+2y = 2e^{3t}+2e^{-t}+4e^{3t}-4e^{-t}=6e^{3t}-2e^{-t}=x′ ✓. y′=6e^{3t}+2e^{-t}; 2x+y=4e^{3t}+4e^{-t}+2e^{3t}-2e^{-t}=6e^{3t}+2e^{-t}=y′ ✓. x(0)=4,y(0)=0 ✓).

Practice Problems

Difficulty 7/10

Find the eigenvalues of A = [[4,1],[2,3]] for the system x′=Ax.

Difficulty 7/10

Convert y″ + 4y = 0 into a first-order system x₁′=?, x₂′=? using x₁=y, x₂=y′.

Difficulty 7/10

A simplified two-species model near equilibrium linearizes to x′=-x+y, y′=x-y (x,y are small population deviations). Find the eigenvalues and describe long-term behavior.

Quiz

Any n-th order linear ODE can be rewritten as:
For x′=Ax, if A has eigenvalue λ with eigenvector v, then a solution of the system is:
In a 2×2 system with two distinct real eigenvalues, the general solution is:

Summary

  • A linear system x′=Ax generalizes scalar ODEs; any higher-order ODE converts into an equivalent first-order system.
  • Solutions are built from eigenvalues and eigenvectors of A: each pair contributes an exponential mode e^{λt}v.
  • The general solution is a linear combination of these independent modes, mirroring the characteristic-root method for scalar equations.

References

  1. WebsiteWikipedia — System of differential equations